I have gone through a proof of the Schröder–Bernstein theorem and would like to know if there are any errors present.
Let $A$ and $B$ be sets and let $f:A\rightarrow B$ and $g:B\rightarrow A$ be injective functions. Define $A_{0}=A$ and $B_{0}=B$. Now define $A_{n}=A-g\left(B-B_{n}\right)$ and $B_{n}=f\left(A_{n-1}\right)$ for all $n\in\mathbb{N}$. It can be shown by induction that $A_{n}\subseteq A_{n-1}$ and $B_{n}\subseteq B_{n-1}$ for all $n\in\mathbb{N}$. Define $X=\bigcap_{n=1}^{\infty}A_{n}$ and $Y=\bigcap_{n=1}^{\infty}B_{n}$.
We now show that the function $f':X\rightarrow Y$ defined by $f'\left(x\right)=f\left(x\right)$ is a bijection.
If $x\in X$, then $x\in A_{n}$ for all $n\in\mathbb{N}$. We then have $f\left(x\right)\in B_{n}$ for all $n\in\mathbb{N}$ and hence $f'\left(x\right)\in Y$. This shows that $f'$ is well defined. Since $f$ is injective, it immediately follows that $f'$ is injective. Let $y\in Y$. Then $y\in B_{n}=f\left(A_{n-1}\right)$ for all $n\in\mathbb{N}$. This means that there exists an element $x_{n}\in A_{n}$ such that $f\left(x_{n}\right)=y$ for all $n\in\mathbb{N}$. Since $f$ is injective, all the $x_{n}$'s are equal. It follows that there exists an element $x\in X$ such that $f'\left(x\right)=y$. Therefore, $f'$ is surjective.
Now we show that the function $g':B-Y\rightarrow A-X$ defined by $g'\left(y\right)=g\left(y\right)$ is a bijection.
If $y\in B-Y$, then there exists an integer $n\in\mathbb{N}$ such that $y\notin B_{n}$. This implies that $g'\left(y\right)\notin A_{n}$. It follows that $g'\left(y\right)\in A-X$. This shows that $g'$ is well defined. Since $g$ is injective, it immediately follows that $g'$ is injective. Let $x\in A-X$. Then there exists an integer $n\in\mathbb{N}$ such that $x\notin A_{n}$. Hence, $x=g'\left(b\right)$ for some $b\in B-B_{n}$. Note that $b\in B-Y$ since $B-B_{n}\subseteq B-Y$.
Now define a function $h:A\rightarrow B$ by $h\left(x\right)=\begin{cases} f'\left(x\right) & \text{if }x\in X,\\ \left(g'\right)^{-1}\left(x\right) & \text{if }x\in A-X. \end{cases}$
The two preceeding arguments show that $h$ is a bijection.
